Question

obtain the particular solution satisfying the initial condition indicated \[y' = x \text{exp} (y - x^2)\] when x = 0; y = 0

steps please?

Answer 1

let's try using variable separation ...
\[ e^{y - x^2} = \frac{e^y}{e^{x^2}}\]

Answer 2

where did e^y and e^x^2 come from?

Answer 3

\[ \huge \frac{dy}{dx} = \frac{x e^y}{e^{x^2}}\]
\[ \huge \int \frac{dy}{e^y} = \int \frac{x}{e^{x^2}} dx\]

Answer 4

x^(a+b) = x^a * x^b

Answer 5

i see..i think that will be all thanks

Answer 6

yw